3.13.32 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{c-a^2 c x^2} \, dx\) [1232]

Optimal. Leaf size=15 \[ -\frac {1}{a c (1+a x)} \]

[Out]

-1/a/c/(a*x+1)

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Rubi [A]
time = 0.03, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6275, 32} \begin {gather*} -\frac {1}{a c (a x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)),x]

[Out]

-(1/(a*c*(1 + a*x)))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6275

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])

Rubi steps

\begin {align*} \int \frac {e^{-2 \tanh ^{-1}(a x)}}{c-a^2 c x^2} \, dx &=\frac {\int \frac {1}{(1+a x)^2} \, dx}{c}\\ &=-\frac {1}{a c (1+a x)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.
time = 0.02, size = 18, normalized size = 1.20 \begin {gather*} -\frac {e^{-2 \tanh ^{-1}(a x)}}{2 a c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)),x]

[Out]

-1/2*1/(a*c*E^(2*ArcTanh[a*x]))

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Maple [A]
time = 0.06, size = 16, normalized size = 1.07

method	result	size
gosper	\(-\frac {1}{a c \left (a x +1\right )}\)	\(16\)
default	\(-\frac {1}{a c \left (a x +1\right )}\)	\(16\)
risch	\(-\frac {1}{a c \left (a x +1\right )}\)	\(16\)
norman	\(\frac {\frac {x}{c}+\frac {a \,x^{2}}{c}}{\left (a x +1\right )^{2}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c),x,method=_RETURNVERBOSE)

[Out]

-1/a/c/(a*x+1)

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Maxima [A]
time = 0.26, size = 14, normalized size = 0.93 \begin {gather*} -\frac {1}{a^{2} c x + a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c),x, algorithm="maxima")

[Out]

-1/(a^2*c*x + a*c)

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Fricas [A]
time = 0.32, size = 14, normalized size = 0.93 \begin {gather*} -\frac {1}{a^{2} c x + a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

-1/(a^2*c*x + a*c)

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Sympy [A]
time = 0.06, size = 12, normalized size = 0.80 \begin {gather*} - \frac {1}{a^{2} c x + a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)**2*(-a**2*x**2+1)/(-a**2*c*x**2+c),x)

[Out]

-1/(a**2*c*x + a*c)

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Giac [A]
time = 0.42, size = 15, normalized size = 1.00 \begin {gather*} -\frac {1}{{\left (a x + 1\right )} a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

-1/((a*x + 1)*a*c)

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Mupad [B]
time = 0.90, size = 13, normalized size = 0.87 \begin {gather*} -\frac {1}{a\,\left (c+a\,c\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a^2*x^2 - 1)/((c - a^2*c*x^2)*(a*x + 1)^2),x)

[Out]

-1/(a*(c + a*c*x))

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